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a) (x^2+2xy+y^2)-9=(x+y)^2-9=(x+y-3)(x+y+3)
b) 5(x^2-2xy+y^2-4z^2)=5[(x-y)^2-4z^2]=5[(x-y-2z)(x-y+2z)
c)x^2-2x-5x+10=x(x-2)-5(x-2)=(x-5)(x-2)
d)2x^2-4x-3x+6=2x(x-2)-3(x-2)=(2x-3)(x-2)
5x2-10xy+5y2-20z2
=5(x2-2xy+y2-4z2)
=5[(x-y)2-4z2]
=5(x-y-2z)(x-y+2z)
dễ hiểu thôi.hơi vắn tắt
a.5x2-10xy+5y2-20z2
=5(x2-2xy+y2-4z2)
=5[ (x2-2xy+y2)-(2z)2 ]
=5[ (x-y)2-(2z)2 ]
=5(x-y-2z)(x-y+2z)
b.16x-5x2-3
=15x+x-5x2-3
=(15x-3)+(x-5x2)
=3(5x-1)+x(1-5x)
=3(5x-1)-x(5x-1)
=(5x-1)(3-x)
c.x2-5x+5y-y2
=(5y-5x)+(x2-y2)
=5(y-x)+(x-y)(x+y)
=5(y-x)-(y-x)(y+x)
=(y-x)[5-(y+x)]
=(y-x)(5-y-x)
d.3x2-6xy+3y2-12z2 (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)
=3(x2-2xy+y2-4z2)
=3[ (x2-2xy+y2)-(2z)2 ]
=3[ (x-y)2-(2z)2 ]
=3(x-y-2z)(x-y+2z)
e.x2+4x+3
=x2+3x+x+3
=(x2+x)+(3x+3)
=x(x+1)+3(x+1)
=(x+1)(x+3)
f.(x2+1)2-4x2
=(x2+1)2-(2x)2
=(x2+1-2x)(x2+1+2x)
h.x2-4x-5
=x2-5x+x-5
=(x2+x)+(-5x-5)
=x(x+1)-5(x+1)
-(x+1)(x-5)
5x2 - 10xy + 5y2 - 20z2
= 5.(x2 - 2xy + y2 - 4z2)
= 5.[(x2 - 2xy + y2) - (2z)2]
= 5.[(x - y)2 - (2z)2]
= 5.(x - y - 2z).(x - y + 2z)
x2.(1 - x2) - 4 + 4x2
= x2.(1 - x2) - 4.(1 - x2)
= (1 - x2).(x2 - 4)
= (1 - x)(1 + x)(x - 2)(x + 2)
5x2 - 10xy + 5y2 - 20z2
= 5.(x2 - 2xy + y2 - 4z2)
= 5.[(x2 - 2xy + y2) - (2z)2]
= 5.[(x - y)2 - (2z)2]
= 5.(x - y - 2z).(x - y + 2z)
x2.(1 - x2) - 4 + 4x2
= x2.(1 - x2) - 4.(1 - x2)
= (1 - x2).(x2 - 4)
= (1 - x)(1 + x)(x - 2)(x + 2)
\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)
\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
a.16x-5x2-3 = - ( 5x2-16x+3) = -( 5x2-15x-x+3)= -[ 5x(x-3)-(x-3)] = -(5x-1)(x-3)
b.x^3-x+3x^2y+3xy^2+y^3-y = \(\left(x^3+3x^2y+3xy^2+y^3\right)-\)\(\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\)\(\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
c.x^4+8x = \(x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
d.x^2+x-6 = \(x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)\)
\(=\left(x+3\right)\left(x-2\right)\)
e.5x^2-10xy+5y^2-20z^2\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)
f.2(x^5)-x^2-5x ( mik ko bik làm)
g.x^3-3x^2-4x+12 = \(x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-2^2\right)\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
h.x^4-5x^2+4 \(=\left(x^2\right)^2-4x^2+4-x^2\)
\(=\left(x^2-2\right)-x^2=\left(x^2-2+x\right)\left(x^2-2-x\right)\)
a/ (5x^2 -10xy+5y^2)-20z^2=5[(x-y)^2-(2z)^2]=5(x-y-2x)(x-y+2z)
b/16x-5x^2-3=-[5x^2-16x+3]=-[(5x^2-x)-(15x+3)]=-[x(5x-1)-3(5x-1)]=(3-x)(5x-1)
c/x^2+4x+3=(x^2+x)+(3x+3)=x(x+1)+3(x+1)=(x+1)(x+3)
2a/ 5x(X^2-9)=0=>x=0 hoặc x^2=9=>x=0 hoặc x=+-3
b/x^2-7x+10=0=>(x^2-2x)-(5x-10)=0=>x(x-2)-5(x-2)=0=>x-2=0 hoặc x-5 =0 => tự tính nhé!
Answer:
Bài 1:
\(5x² - 10xy + 5y² - 20z²\)
\(= 5( x² - 2xy + y² - 4z²)\)
\(= 5 [(x² - 2xy + y²) - (2z)²]\)
\(= 5 [(x - y)² - (2z)²]\)
\(= 5 (x - y - 2z) ( x - y + 2z)\)
\(16x - 5x² - 3 \)
\(= -( 5x² - 16x + 3)\)
\(= -( 5x² - 15x - 1x + 3)\)
\(= - [ (5x² -x) - (15x -3)]\)
\(= - [ x(5x -1) -3(5x -1)]\)
\(= - (5x-1)(x-3)\)
\(x² + 4x + 3\)
\(= x² + x + 3x + 3\)
\(= (x² + x) + (3x + 3)\)
\(= x( x + 1) +3 (x+1)\)
\(= (x+1) (x+3)\)
Bài 2:
\(5x\left(x^2-9\right)=0\)
\(\Rightarrow5x\left(x-3\right)\left(x+3\right)=0\)
Trường hợp 1: \(5x=0\Leftrightarrow x=0\)
Trường hợp 2: \(x-3=0\Leftrightarrow x=3\)
Trường hợp 3: \(x+3=0\Leftrightarrow x=-3\)
\(x^2-7x+10=0\)
\(\Rightarrow x^2-5x-2x+10=0\)
\(\Rightarrow x\left(x-5\right)-2\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}}\)